Question

The length of the latus-rectum of the ellipse, whose foci are $(2, 5)$ and $(2, -3)$ and eccentricity is $\frac{4}{5}$, is

• A. $\frac{6}{5}$
• B. $\frac{50}{3}$
• C. $\frac{10}{3}$
• D. $\frac{18}{5}$

Hint:

Use the relationship between the semi-major axis, semi-minor axis, and the distance between the foci to find the length of the latus-rectum.

Answer 1

The Correct Option is D

1. Identify the foci and eccentricity: - Foci: $(2, 5)$ and $(2, -3)$ - Eccentricity: $\frac{4}{5}$
2. Calculate the distance between the foci: \[ 2c = |5 - (-3)| = 8 \implies c = 4 \]
3. Use the relationship between $a$, $b$, and $c$: \[ e = \frac{c}{a} = \frac{4}{5} \implies a = 5 \] \[ b^2 = a^2 - c^2 = 25 - 16 = 9 \implies b = 3 \]
4. Calculate the length of the latus-rectum: \[ \text{Length of latus-rectum} = \frac{2b^2}{a} = \frac{2 \cdot 3^2}{5} = \frac{18}{5} \] Therefore, the correct answer is (4) $\frac{18}{5}$.

Answer 2

Given foci of an ellipse at \((2,5)\) and \((2,-3)\) and eccentricity \(e=\frac{4}{5}\).

Concept Used:

For an ellipse with semi-major axis \(a\), semi-minor axis \(b\), and focal distance \(c\), we have \[ e=\frac{c}{a},\qquad b^2=a^2-c^2, \] and the length of the latus-rectum is \[ \text{LR}=\frac{2b^2}{a}. \]

Step-by-Step Solution:

Step 1: Find \(c\).

The center is the midpoint of the foci: \((2,1)\). The distance from the center to a focus is \[ c=\sqrt{(2-2)^2+(5-1)^2}=4. \]

Step 2: Use eccentricity to get \(a\).

\[ e=\frac{c}{a}=\frac{4}{a}=\frac{4}{5}\ \Rightarrow\ a=5. \]

 

Step 3: Compute \(b^2\).

\[ b^2=a^2-c^2=25-16=9. \]

 

Step 4: Length of latus-rectum.

\[ \text{LR}=\frac{2b^2}{a}=\frac{2\cdot 9}{5}=\frac{18}{5}. \]

 

Final Computation & Result

Length of latus-rectum = \(\dfrac{18}{5}\).